hw fem.pdf

8/11/2019 HW fem.pdf

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Code used for the exercise is 135

According to the code given, the structure that will be analyzed is:

Its a beam structure with a circular cross section that are loaded by a distributed force, p, aconcentrated force, F1, and a concentrated moment, M1. The material of the beams is linear

elastic, homogeneous and isotropic with na elastic modulus of E and v=0,3. For this case it

have to be considered that l[m]= 1.1, d1[m]= 0.070, p[N/m]= 470, M1[Nm]= 350, d2[m]= 0.105,

F1[N]= 820 and E[GPa]= 206.

1. Calculating the displacement and angle of rotation function using the 0th-

8th approximation:

Is necessary to use the total potential energy:

Where:

The kinematic Boundary conditions in this case that is satisfied to the equation is:


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And:

After that the equation for the Total Potential Energy is of the beam structure is:

Where, according to the Ritz method the Total energy has to be minimal:

Using the Mathematica 9.0 was possible to obtain the results. The code is

showed below:

l=1.1;d1=0.070;p=470;M=350;d2=0.105;F=820;Ey=206*10^9;Iz1=(3.14*(d1/2)^4)/4;

Iz2=(3.14*(d2/2)^4)/4;

omega=(x^2)*(x-5*l);Nappr=0;

B=omega*Table[{x^(i-1)},{i,1,Nappr+1}];Bt=Transpose[B];dB=D[B,x];ddB=D[B,{x,2}];ddBt=Transpose[ddB];

S=Integrate[Iz2*Ey*ddB.ddBt,{x,0,3*l}]+Integrate[Iz1*Ey*ddB.ddBt,{x,3*l,5*l}];

Q=Integrate[-p*B,{x,0,3*l}]-(F*B/.x 4*l)+(M*dB/.x 3*l);A=(1/S).Q;

v[x_]=Bt.A;phi[x_]=D[v[x],x];vK=v[3*l]phiK=phi[3*l]


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The results is showed in the table:

Numbers of approximation(N1)

Displacement (v) Angle of Rotation

0th -0,00292942 -0,0004438521th -0,00906533 -0,00250898

2th 0,0147005 -0,00540155

3th -0,0189595 -0,0081995

4th -0,0220144 -0,0105235

5th -0,0241998 -0,0123216

6th -0,025787 -0,0136653

7th -0,0269624 -0,0146527

8th -0,0278491 -0,0153729

And represented by the diagram:

= (-0,0153729 rad/ -

2- The FE model of the structure using the Mathematica, applying the

minimum number of elements:

The number of elements in this case is 3. The element applied is BEAM1D. The

global system is enough for this exercise because the beam is parallel with the axis x

and perpendicular to the plane xz. The solution was found by the Mathematica 9.0,

according to this code:

-0,00292942

-0,00906533

-0,0147005

-0,0189595

-0,0220144

-0,0241998-0,025787

-0,0269624-0,0278491

-0,03

-0,025

-0,02

-0,015

-0,01

-0,005

0

0 2 4 6 8 10

vkx N

Displacement


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l=1.1;d1=0.07;p=470;d2=0.105;Ey=206*10^9;Iz1=((d1^4)*Pi)/64;Iz2=((d2^4)*Pi)/64;l3=3*l;v=0.3;

N1={{6,l3,-6,l3},{l3,2*l^2,-l3,l^2},{-6,-l3,6,-l3},{l3,l^2,-l3,2*l^2}};N2={{6,l3,-6,l3},{l3,2*l^2,-l3,l^2},{-6,-l3,6,-l3},{l3,l^2,-l3,2*l^2}};N3={{6,3*l3,-6,3*l3},{3*l3,2*l3^2,-3*l3,l3^2},{-6,-3*l3,6,-3*l3},{3*l3,l3^2,-3*l3,2*l3^2}};

Ke3=(2*Iz1*Ey/l^3)*N1;Ke2=(2*Iz1*Ey/l^3)*N2;Ke1=(2*Iz2*Ey/l3^3)*N3;

U={{0,0,v2,fi2,v3,fi3,0,fi4}};Ut=Transpose[U];

Ff={{Ay,Ma,0,220,-820,0,By,0}};Fp=(p*l3/12)*{{6,l3,6,-l3,0,0,0,0}};F=Ff-Fp;Ft=Transpose[F];

H1={{0,0,0,0},{0,0,0,0},{0,0,0,0},{0,0,0,0}};H2={{0,0,0,0,0,0,0,0},{0,0,0,0,0,0,0,0}};H3={{0,0},{0,0},{0,0},{0,0}};

K1=ArrayFlatten[{{Ke1,H1},{H1,H1}}];K2A=ArrayFlatten[{{H3,Ke2,H3}}];

K2=ArrayFlatten[{{H2},{K2A},{H2}}];K3=ArrayFlatten[{{H1,H1},{H1,Ke3}}];K=K1+K2+K3;

adof={3,4,5,6,8};Kcon=K[[adof,adof]];Fcon=Ft[[adof]];Ucon=LinearSolve[Kcon,Fcon];Ut[[adof]]=Ucon;

fixdof={1,2,7};sol=Solve[(K.Ut)[[fixdof]] Ft[[fixdof]],{Ay,Ma,By}]


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According to the code was possible to find the Structural Stiffness matrix, that is:

The load vector is:

The vector of nodal displacements:

Solving the FE problem for this data we have this results:

Node 1: v(1)= 0[m]; fi(1)=0[rad];Node 2: v(2)= [m]; fi(2)= [rad];Node 3: v(3)= [m]; fi(3)= [rad];Node 4: v(3)= 0[m]; fi(4)= [rad];

Reactions: F1=[N]; F4=[N]; M1=[Nm];

3- The FE model of the structure using Ftool is:The FE model has 3 elements for the same reasons for the Mathematica solution insecond question. For this exercise we can get the results after define the nodes,elements, constraints, types of materials & of cross-sections and loads. We have allthe necessary data, therefore, the diagram plotted are:

The Structure:


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Deflection diagram:

The bending moment diagram:

Shear force diagram:


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Elemento 1 Elemento 2 Elemento 3


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Node 2 Node 3 Node 1 Node 4

a ra is:

= (-.754/ -.45

Conclusion:

After all the calculation and the use of different methods we could observe that the results ofthe Ftool method was the best, showing a good and precise result. So, to aproximate the Ritzmethod to the Ftool and the Mathematica, is necessary to use more numbers ofapproximation.

hw fem.pdf

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